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A number is the reciprocal of the other. If the arithmetic mean of the two numbers be $\frac{13}{12}$, then the numbers are
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Verified Answer
The correct answer is:
$\frac{3}{2}, \frac{2}{3}$
Suppose that required numbers $a$ and $b$ Therefore according to the conditions $a=\frac{1}{b}$
and $\frac{a+b}{2}=\frac{13}{12} \Rightarrow$ $a+b=\frac{13}{6}$
$\Rightarrow \quad a+\frac{1}{a}=\frac{13}{6}$ $\Rightarrow 6 a^2-13 a+6=0$
$\Rightarrow\left(a-\frac{3}{2}\right)\left(a-\frac{2}{3}\right)=0$
$\Rightarrow \quad a=\frac{3}{2}$ and $b=\frac{2}{3}$
or $a=\frac{2}{3}$ and $b=\frac{3}{2}$
and $\frac{a+b}{2}=\frac{13}{12} \Rightarrow$ $a+b=\frac{13}{6}$
$\Rightarrow \quad a+\frac{1}{a}=\frac{13}{6}$ $\Rightarrow 6 a^2-13 a+6=0$
$\Rightarrow\left(a-\frac{3}{2}\right)\left(a-\frac{2}{3}\right)=0$
$\Rightarrow \quad a=\frac{3}{2}$ and $b=\frac{2}{3}$
or $a=\frac{2}{3}$ and $b=\frac{3}{2}$
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